Simplify; express your answer in exponential form. Assume $n\neq 0, p\neq 0$. $\dfrac{{n^{3}p^{-1}}}{{(n^{-2}p^{-2})^{-4}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${n^{3}p^{-1} = n^{3}p^{-1}}$ On the left, we have ${n^{3}}$ to the exponent ${1}$ . Now ${3 \times 1 = 3}$ , so ${n^{3} = n^{3}}$ Apply the ideas above to simplify the equation. $\dfrac{{n^{3}p^{-1}}}{{(n^{-2}p^{-2})^{-4}}} = \dfrac{{n^{3}p^{-1}}}{{n^{8}p^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{3}p^{-1}}}{{n^{8}p^{8}}} = \dfrac{{n^{3}}}{{n^{8}}} \cdot \dfrac{{p^{-1}}}{{p^{8}}} = n^{{3} - {8}} \cdot p^{{-1} - {8}} = n^{-5}p^{-9}$